Prob. 11, Chap. 4 in Baby Rudin: uniformly continuous extension from a dense subset to the entire space

Here is Prob. 11, Chap. 4 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Suppose $f$ is a uniformly continuous mapping of a metric space $X$ into a metric space $Y$ and prove that $\left\{ f\left( x_n \right) \right\}$ is a Cauchy sequence in $Y$ for every Cauchy sequence $\left\{ x_n \right\}$ in $X$. Use this result to give an alternative proof of the theorem stated in Exercise 13.

Now here is Prob. 13, Chap. 4 in Baby Rudin, 3rd edition:

Let $E$ be a dense subset of a metric space $X$, and let $f$ be a unifromly continuous real function defined on $E$. Prove that $f$ has a continuous extension from $E$ to $X$. ... (Uniqueness follows from Exercise 4.) ...

And, here is Prob. 4, Chap. 4:

Let $f$ and $g$ be continuous mappings of a metric space $X$ into a metric space $Y$, and let $E$ be a dense subset of $X$. Prove that $f(E)$ is dense in $f(X)$. If $g(p) = f(p)$ for all $p \in E$, prove that $g(p) = f(p)$ for all $p \in X$. (In other words, a continuous mapping is determined by its values on a dense subset of its domain.)

My effort:

Since $f$ is uniformly continuous on $X$, corresponding to every real number $\varepsilon > 0$, we can find a real number $\delta > 0$ such that $$d_Y\left(f(x), f(y)\right) < \varepsilon$$ for all points $x, y \in X$ which satisfy $$d_X(x,y)<\delta.$$

Now since $\left\{ x_n \right\}$ is a Cauchy sequence in $\left( X, d_X \right)$, therefore corresponding to the real number $\delta$ in the preceding paragraph we can find a natural number $N$ such that $$d_X\left(x_m, x_n \right)< \delta \ \mbox{ for any natural numbers } m \mbox{ and } n \mbox{ such that } m > N \mbox{ and } n > N.$$

So from the preceding two paragraphs we can conclude that $$d_Y\left( f\left(x_m\right), f\left(x_n\right) \right) < \varepsilon \ \mbox{ for any natural numbers } m \mbox{ and } n \mbox{ such that } m > N \mbox{ and } n > N,$$ from which it follows that $\left\{ f\left(x_n\right) \right\}$ is a Cauchy sequence in $\left( Y, d_Y \right)$.

Now let $E$ be a dense set in $\left(X, d_X \right)$, $\left( Y, d_Y \right)$ be a complete metric space, and $f$ be a uniformly continuous mapping of $E$ into $Y$.

We now show that there exists a unique uniformly continuous mapping $g$ of $X$ into $Y$ such that $g(p) = f(p)$ for all $p \in E$.

Let $p \in X$. Since $E$ is dense in $X$, we can find a sequence $\left\{p_n\right\}$ in $E$ converging to the point $p$ in $\left( X, d_X \right)$.

Now as the sequence $\left\{ p_n \right\}$ is a Cauchy sequence in $E$ and as $f$ is uniformly continuous, so the image sequence $\left\{ f\left(p_n \right)\right\}$ is a Cauchy sequence in the complete metric space $Y$, and so this sequence converges to some point $q$ in $Y$.

We now show that this point $q$ is independent of the particular sequence $\left\{ p_n \right\}$ in $E$ which converges to $p$. For this, let $\left\{ p^\prime_n \right\}$ be another sequence in $E$ converging in $X$ to the point $p$. Then as before the image sequence $\left\{ f\left(p^\prime_n\right)\right\}$ converges in $Y$ to some point $q^\prime$. Now consider the sequence $\left\{ x_n \right\}$ in $E$ defined as follows: Let $$x_n = \begin{cases} p_{\frac{n}{2}} \ \mbox{ if n is even}; \\ p^\prime_{\frac{n+1}{2}} \ \mbox{ if n is odd}. \end{cases}$$

This sequence too converges to point $p$ in $X$ and is therefore Cauchy and so its image sequence $\left\{ f\left( x_n \right) \right\}$ is also a Cauchy sequence in $Y$, which is a complete metric space; so the sequence $\left\{ f\left( x_n \right) \right\}$ converges in $Y$ to some point $y$. Therefore every subsequence of $\left\{ f\left( x_n \right) \right\}$ also converges to $y$.

But we note that, $$p_n = x_{2n} \ \mbox{ and } p^\prime_n = x_{2n-1} \ \mbox{ for all } n \in \mathbb{N}.$$ So $$q = \lim_{n \to \infty} f\left(p_n \right) = \lim_{n \to \infty} f\left( x_{2n}\right) = y,$$ where the limit is in $Y$, and similarly $$q^\prime = \lim_{n\to\infty} f\left(p_{2n-1}\right) = y.$$ Thus $q^\prime = q$.

Now let's define the mapping $g$ of $X$ into $Y$ as follows: For any point $p \in X$, let $$\tag{1} g(p) = \lim_{n \to \infty} f \left(p_n \right),$$ the limit being in the metric space $\left(Y, d_Y \right)$, where $\left\{ p_n \right\}$ is any sequence in $E$ such that $$\tag{2} \lim_{n \to \infty} p_n = p$$ in $\left( X, d_X \right)$.

If $p \in E$, then we can take $p_n = p$ for all $n$ so that $$g(p) = \lim_{n \to \infty} f\left(p_n\right) = f(p).$$ Thus $g$ is an extension of $f$ from $E$ to $X$.

Let $\varepsilon > 0$ be a given real number. Since $f$ is uniformly continuous on $E$, we can find a real number $\delta > 0$ such that $$d_Y\left( f(p), f(q) \right) < \frac{\varepsilon}{3}$$ for all points $p$ and $q$ in $E$ for which $$d_X\left(p, q\right) < \delta.$$

Now let $u$ and $v$ be any two points of $X$ which satisfy $$d_X\left(u, v\right) < \frac{\delta}{3}.$$ Then since $E$ is dense in $X$, we can find points $u^\prime$ and $v^\prime$ in $E$ such that $$\tag{3} d_X \left( u, u^\prime \right) < \frac{\delta}{3} \mbox{ and } d_Y\left(g(u), f(u^\prime) \right) < \frac{\varepsilon}{3},$$ and also $$\tag{4} d_X \left( v, v^\prime \right) < \frac{\delta}{3} \mbox{ and } d_Y\left( g(v), f(v^\prime) \right) < \frac{\varepsilon}{3}.$$ This is possible in view of (1) and (2) above.

Then we note that $$d_X \left( u^\prime, v^\prime \right) \leq d_X\left(u^\prime, u \right) + d_X\left(u, v\right) + d_X\left(v, v^\prime\right) < \frac{\delta}{3} + \frac{\delta}{3} + \frac{\delta}{3} = \delta,$$ which implies that $$\tag{5} d_Y\left( f\left(u^\prime \right), f\left(v^\prime\right) \right) < \frac{\varepsilon}{3},$$ and therefore from (3), (4), and (5) we can conclude that \begin{align} d_Y\left( g\left(u\right), g\left(v\right) \right) &\leq d_Y\left( g\left(u \right), f\left(u^\prime \right) \right) + d_Y\left( f\left(u^\prime\right), f\left(v^\prime\right) \right) + d_Y\left( f\left(v^\prime\right), g\left(v\right) \right) \\ &< \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} \\ &= \varepsilon. \end{align}

Let us take a real number $\eta$ such that $$0 < \eta < \frac{\delta}{3}.$$

Thus, corresponding to every real number $\varepsilon > 0$, we can find a positive real number $\eta$ such that $$d_Y \left( g(u), g(v) \right) < \varepsilon$$ for all points $u$ and $v$ in $X$ for which $$d_X \left(u, v \right) < \eta.$$ Hence $g$ is uniformly continuous on $X$.

Let $g^\prime$ be (also) a (uniformly) continuous extension of $f$ from $E$ to $X$.

Since $g^\prime(p) = f(p) = g(p)$ for all $p \in E$ and since $E$ is dense in $X$, therefore by virtue of the conclusion in Prob. 4 we can conclude that $g^\prime(p) = g(p)$ for all $p \in X$.

Now I know that the first part of the above solution, where we are required to show that $\left\{ f\left(x_n\right) \right\}$ is a Cauchy sequence in $Y$ for every Cauchy sequence $\left\{ x_n \right\}$ in $E$, is correct, isn't it?

What about the second part? Is the result I've stated correct? If so, then is my proof correct also?

• Everything seems correct to me. – Crostul Feb 1 '17 at 9:50
• @Crostul Kudos to you for reading the entire post. – zhw. Mar 29 '17 at 22:45
• Re: Prob.4 Chap.4: "In other words a continuous function is completely determined by its values on a dense set." This is because the range space is Hausdorff. Theorem: Let $f:X\to Y$ and $g:X\to Y$ be continuous. If $Y$ is Hausdorff then $\{x\in X: f(x)=g(x)\}$ is closed in $X.$ Equivalently (and easier to prove) $\{x\in X: f(x)\ne g(x)\}$ is open in $X$..... It's not hard. You use the property that if $f(x)\subset V$ where $V$ is open in $Y$ then $x\in U$ for some open $U\subset X$ such that $f(U)\subset V.$ – DanielWainfleet Mar 30 '17 at 0:17